This is a follow-up of this question.

Let $V,W$ be $d$-dimensional real vector spaces, and let $A,B \in \text{Hom}(V,W)$ be non-invertible maps, $1 \le k \le d-1$.

Consider the induced maps $\bigwedge^{k}A,\bigwedge^{k}B :\Lambda_k(V) \to \Lambda_k(W)$. I want to characterize all the pairs $(A,B)$ which satisfy $\bigwedge^k A=\bigwedge^k B \neq 0.$

Partial results:

$(1)$ I prove below that a necessary condition is $\ker A=\ker B, \text{Image} \, A=\text{Image} \, B$. In particular $\text{rank} \,A=\text{rank} \,B$.

$(2)$ Moreover, contrary to the case where $A,B$ are invertible when the only possible pairs are $(A,A),(A,-A)$, in the singular case there is much more freedom:

Take a $3$-dimensional space with a basis $v_1,v_2,v_3$, and define $A,B$ via: $Av_3=Bv_3=0,Av_1=2v_1,Av_2=\frac{1}{2}v_2,Bv_1=\frac{1}{2}v_1,Bv_2=2v_2$.

This is can be generalized into the following sufficient condition:

$A,B$ have identical eigenspaces, null spaces, and equal products of all the nonzero eigenvalues.

Essentially, since $\ker A=\ker B=W$, $\text{Image} \, A=\text{Image} \, B=\tilde W$, the question really seems about the invertible quotient operators:

$\tilde A,\tilde B:V/W \to \tilde W$.

(I think my example above is related to the special case where $k=\text{rank} \,B=\dim (\tilde W)$. The right condition thenseems to be the "determinant" of the quotient operator is $1$).

Proof that $\ker A=\ker B$:

Let $v \in \ker A$ and assume by contradiction that $Bv \neq 0$. Fix some inner product on $V$, in such a way that $v \in (\ker B)^{\perp}$. Since $\dim(\ker B)^{\perp} = \text{rank} {B} \ge k$, there exist $v_1,\dots,v_{k-1} \in (\ker B)^{\perp}$ such that $v,v_1,\dots,v_{k-1}$ are linearly independent.

Then $ 0=\bigwedge^k A(v \wedge v_1 \wedge \dots \wedge v_{k-1})= Bv \wedge Bv_1 \wedge \dots \wedge Bv_{k-1}$, hence $Bv,Bv_1,\dots,Bv_{k-1}$ are linearly dependent, in contradiction to the fact that $B|_{(\ker B)^{\perp}}$ is injective.

This shows $\ker A \subseteq \ker B$. The other direction follows from symmetry.

The proof that $\text{Image} \, A=\text{Image} \, B$ is easy.

*(I am sure there is a cleaner argument with no insertion of an inner product, e.g via quotients).

  • $\begingroup$ Are you sure? I meant to assume equality for a single $k$, not all of them... $\endgroup$ – Asaf Shachar Jul 8 '17 at 15:44
  • $\begingroup$ The proof works for a single fixed $k$, at least to me. $\endgroup$ – Gunnar Þór Magnússon Jul 8 '17 at 16:00

It seems to me that all the ingredients of a full answer are included in your question. I'll just put them in the right order.

Yes, $A$ and $B$ have the same kernel $U$ and the same image $\tilde{W}$, so we may think of the two invertible operators $$A',B':V/U\to\tilde{W}.$$

Now, you just need to distinguish between two cases. If $\dim\tilde{W}=k$, then a necessary and sufficient condition is that both operators have the same determinant (this is not even a condition, just another way to formulate the equality you are talking about).

If $\dim\tilde{W}>k,$ the answer to your previous question applies again. That is, $A'$ and $B'$ have to differ by a constant $\alpha$ satisfying $\alpha^k=1$.

It is interesting that the first case allows a big space of solutions (something like $\text{SL}(V/U))$, whereas the solutions in the second case are all trivial.

  • $\begingroup$ Thanks. Somehow I feel that the notion of determinant of the quotient maps $A',B'$ shouldn't be well-defined, since in general the determinant of a linear map between two different vector spaces, is not well-defined (unless you add some additional structure such as metrics or preferred volume forms on these spaces). This was the reason for my hesitation in putting "determinant". For a map $V \to V$ (that is from a vector space to itself) the determinant is always defined without any need for more structure (just take some top-form $\omega$, $\endgroup$ – Asaf Shachar Jul 8 '17 at 20:45
  • $\begingroup$ then $\det T$ is defined by $T^*\omega =\det T \cdot \omega$. $\endgroup$ – Asaf Shachar Jul 8 '17 at 20:47
  • $\begingroup$ Nevertheless, @AsafShachar, saying that two different operators $V\to W$ have the same determinant is well defined, even with no extra structure. If you like, you can replace this by "$\det B^{-1}A=1$". $\endgroup$ – Amitai Yuval Jul 8 '17 at 21:23
  • $\begingroup$ @AsafShachar In other words, this just depends on your definition for the determinant of a linear map $V\to W$. My definition is the induced linear map of the top exterior powers. If $V=W$, this map can be thought of as a number. $\endgroup$ – Amitai Yuval Jul 8 '17 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.